Problem: There is a light source on the plane. This source is so small that it can be represented as point. The light source is moving from point (a, s_y) to the (b, s_y) (s_y < 0) with speed equal to 1 unit per second. The trajectory of this light source is a straight segment connecting these two points. 

There is also a fence on OX axis represented as n segments (l_i, r_i) (so the actual coordinates of endpoints of each segment are (l_i, 0) and (r_i, 0)). The point (x, y) is in the shade if segment connecting (x,y) and the current position of the light source intersects or touches with any segment of the fence.

<image>

You are given q points. For each point calculate total time of this point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y).

Input

First line contains three space separated integers s_y, a and b (-10^9 ≤ s_y < 0, 1 ≤ a < b ≤ 10^9) — corresponding coordinates of the light source.

Second line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of segments in the fence.

Next n lines contain two integers per line: l_i and r_i (1 ≤ l_i < r_i ≤ 10^9, r_{i - 1} < l_i) — segments in the fence in increasing order. Segments don't intersect or touch each other.

Next line contains single integer q (1 ≤ q ≤ 2 ⋅ 10^5) — number of points to check.

Next q lines contain two integers per line: x_i and y_i (1 ≤ x_i, y_i ≤ 10^9) — points to process.

Output

Print q lines. The i-th line should contain one real number — total time of the i-th point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y). The answer is considered as correct if its absolute of relative error doesn't exceed 10^{-6}.

Example

Input

-3 1 6
2
2 4
6 7
5
3 1
1 3
6 1
6 4
7 6


Output

5.000000000000000
3.000000000000000
0.000000000000000
1.500000000000000
2.000000000000000

Note

  * The 1-st point is always in the shade; 
  * the 2-nd point is in the shade while light source is moving from (3, -3) to (6, -3); 
  * the 3-rd point is in the shade while light source is at point (6, -3). 
  * the 4-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and at point (6, -3); 
  * the 5-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and from (5.5, -3) to (6, -3); 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <class TH>
void _dbg(const char *sdbg, TH h) {
  cerr << sdbg << " = " << h << endl;
}
template <class TH, class... TA>
void _dbg(const char *sdbg, TH h, TA... a) {
  while (*sdbg != ',') cerr << *sdbg++;
  cerr << " = " << h << ", ";
  _dbg(sdbg + 1, a...);
}
const int maxn = (1e6) + 7;
const int maxk = 20;
const int inf = (1e9) + 7;
const long long LLinf = (1e18) + 7;
const double eps = 1e-9;
const long long mod = 1e9 + 7;
bool gora(complex<double> a) {
  if (a.imag() > eps || (a.imag() >= eps && a.real() >= eps)) return 1;
  return 0;
}
long long wyz(complex<double> a, complex<double> b, complex<double> c) {
  b -= a;
  c -= a;
  return b.real() * c.imag() - b.imag() * c.real();
}
int wyzznak(complex<double> a, complex<double> b, complex<double> c) {
  long long res = wyz(a, b, c);
  if (res > 0) return 1;
  if (res == 0) return 0;
  return -1;
}
long long skalar(complex<double> a, complex<double> b) {
  return a.real() * b.real() + a.imag() * b.imag();
}
bool angcomp(complex<double> &a, complex<double> &b) {
  if (gora(a) != gora(b)) return gora(b) < gora(a);
  return wyzznak(complex<double>(0, 0), a, b) == 1;
}
struct line {
  double a, b, c;
  line(double A, double B, double C) {
    a = A;
    b = B;
    c = C;
  }
  line(complex<double> A, complex<double> B) {
    a = A.imag() - B.imag();
    b = B.real() - A.real();
    c = A.imag() * (A.real() - B.real()) + (B.imag() - A.imag()) * A.real();
  }
};
complex<double> inter(line a, line b) {
  if (fabs(a.a * b.b - a.b * b.a) < eps) return {1e18, 1e18};
  return {(b.b * a.c - a.b * b.c) / (b.a * a.b - a.a * b.b),
          (b.a * a.c - a.a * b.c) / (a.a * b.b - b.a * a.b)};
}
int lewo[maxn], prawo[maxn], suma[maxn];
int n;
double ile(double xx) {
  xx = min(1e9, xx);
  int cnt = ceil(xx);
  int l = 0;
  int p = n + 1;
  while (l + 1 < p) {
    int sr = (l + p) / 2;
    if (cnt <= prawo[sr])
      p = sr;
    else
      l = sr;
  }
  xx = max(xx, (double)lewo[p]);
  return suma[p - 1] + xx - lewo[p];
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  double h;
  double l, p;
  cin >> h >> l >> p;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> lewo[i] >> prawo[i];
    suma[i] = suma[i - 1] + prawo[i] - lewo[i];
  }
  suma[n + 1] = suma[n];
  prawo[n + 1] = lewo[n + 1] = inf;
  int m;
  cin >> m;
  complex<double> pkt1(l, h);
  complex<double> pkt2(p, h);
  line xdxd({0.0, 0.0}, {1.0, 0.0});
  while (m--) {
    double a, b;
    cin >> a >> b;
    complex<double> cur(a, b);
    complex<double> xd = inter(xdxd, {cur, pkt1});
    double res = -ile(xd.real());
    xd = inter(xdxd, {cur, pkt2});
    res += ile(xd.real());
    printf("%.9lf\n", (res / cur.imag()) * (cur.imag() - h));
  }
  return 0;
}